Suppose a student started with 129 mg of trans-cinnamic acid and 0.50 mL of a 10% (v/v) bromime solution, and after the reaction and workup, ended up with 0.216 g of brominated product. Calculate the student\'s theoretical and percent yields.
1. Write the equation and balance it.
2. Convert 129 mg to mols. mols = grams/molar mass
3. Convert 0.50 mL of 10% bromine to mols. I believe this is 0.05 mL bromine. You may need to look up the density of liquid bromine to and use that to obtain grams, then convert to mols.
4. Using the coefficients in the balanced equation to convert mols cinnamic acid to mols of the product.
5. Do the same for bromine.
6. The two answers probably will be different; the correct answer in limiting reagent problems (such as this) is ALWAYS the smaller value tna the reagent producing that number is the limiting reagent.
7. Convert mols of the product to grams. This is the theoretical yield (TE).
8. Then %yield = (actual yield/TE)*100 = ?
Note: If you KNOW which is the limiting reagent you can use it for step 1, 2or 3, 4or 5, then 6 and 7.
What does the equation look like? What is the grams?
The lim reg is trans-cinn.
so convert 129 mg trans-cinn to moles using its molar mass (148.16g/mol), then use 1 to 1 mole ratio to get to moles of product, then convert to grams product using molar mass 307.97g/mol. this gives you theo yield.
note: u don't need to use the ml bromine in the calculations, it was only to find which is the lim reagent--which i've given you.
To calculate the student's theoretical and percent yields, we need to understand and use the concepts of theoretical yield and percent yield.
1. Theoretical Yield:
The theoretical yield is the maximum amount of product that can be formed in a chemical reaction, assuming perfect conditions and complete conversion of the limiting reactant.
2. Percent Yield:
The percent yield is the ratio of the actual yield (the experimental amount of product obtained) to the theoretical yield, expressed as a percentage. It indicates the efficiency of the reaction.
To calculate the theoretical and percent yields, we need to follow the following steps:
Step 1: Convert the given quantities to the same unit.
- Convert 0.50 mL of bromine solution to grams (since mass is a commonly used unit).
- Convert 129 mg of trans-cinnamic acid to grams.
Step 2: Determine the limiting reactant.
- Compare the amounts of each reactant, trans-cinnamic acid, and bromine solution, to identify which one is the limiting reactant using stoichiometry.
Step 3: Calculate the theoretical yield.
- Use the balanced chemical equation and the amount of limiting reactant to determine the moles of the desired product.
- Convert moles of product to grams.
Step 4: Calculate the percent yield.
- Divide the actual yield (0.216 g) by the theoretical yield and multiply by 100 to get the percent yield.
Let's go through these steps in detail:
Step 1: Converting the given quantities to the same unit:
- 0.50 mL of bromine solution to grams:
The density of the bromine solution is not provided, so we will assume it to be 1 g/mL (since the density of water is close to 1 g/mL).
Therefore, the mass of 0.50 mL of the bromine solution is 0.50 g.
- 129 mg of trans-cinnamic acid to grams:
Since 1 g = 1000 mg, we can convert 129 mg to grams by dividing it by 1000.
Therefore, 129 mg = 0.129 g.
Step 2: Determine the limiting reactant:
To determine the limiting reactant, we need to compare the mole-to-mole ratio between the reactants and the desired product.
The balanced chemical equation for the reaction is not given, so we cannot determine the exact ratio. We will assume a 1:1 ratio for simplicity.
In this case, both trans-cinnamic acid and bromine solution are in excess. So, there is no limiting reactant, and the theoretical yield will be based on the trans-cinnamic acid.
Step 3: Calculate the theoretical yield:
- Using the assumed 1:1 mole ratio, we can calculate the moles of the brominated product using the given amount of trans-cinnamic acid.
Moles of the brominated product = Mass of trans-cinnamic acid / molar mass of trans-cinnamic acid.
- Calculate the molar mass of trans-cinnamic acid (C9H9COOH):
C: 12.01 g/mol × 9 = 108.09 g/mol
H: 1.01 g/mol × 10 = 10.10 g/mol
O: 16.00 g/mol
Molar mass of trans-cinnamic acid = 108.09 g/mol + 10.10 g/mol + 16.00 g/mol = 134.19 g/mol
- Moles of brominated product = 0.129 g / 134.19 g/mol = 0.000961 mol.
- Convert moles of product to grams:
Mass of brominated product = Moles of brominated product × molar mass of brominated product.
As the molar mass of the brominated product is not provided, we cannot calculate the exact mass. We will need information about the molecular formula or molar mass of the product to proceed.
Step 4: Calculate the percent yield:
- Given actual yield = 0.216 g brominated product.
- Percent yield = (Actual yield / Theoretical yield) × 100.
(Actual yield / Theoretical yield) = (0.216 g / Theoretical yield) × 100.
Without the information regarding the molar mass or molecular formula of the brominated product, we cannot determine the percent yield accurately.
Please provide the molecular formula or molar mass of the brominated product to proceed with the final calculation of the percent yield.